Local Pfaff Geometry — Examples

The first derivatives of the coefficients $2y$ and $1+x$ are $0,2$ and $1,0$, respectively, and all their second partial derivatives vanish; hence $\omega$ is $C^2$ and nowhere zero. Expanding $S(x,y)=(1+x{)}^{2}y$ gives $S(x+h,y+k)-S(x,y)=2y(1+x)h+(1+x{)}^{2}k+2(1+x)hk+y{h}^2+h^{2}k$. The last three terms divided by $\sqrt{h^2+k^2}$ tend to zero as $(h,k)\to (0,0)$, so $d{S}_{(x,y)}(h,k)=2y(1+x)h+(1+x{)}^{2}k$. These first derivative coefficients have continuous first partials $S_{xx}=2y$, $S_{xy}=S_{yx}=2(1+x)$ and $S_{yy}=0$, so $S\in C^2$. Thus $dS=2y(1+x)dx+(1+x{)}^{2}dy=(1+x)\omega =\mu \omega$. Because $1+x>0$ on $U$, both $\mu$ and $dS$ are nowhere zero.

For any piecewise-$C^1$ horizontal path $\gamma$, the identity in step 1.1 and [F1]–[F2] give $(S\circ \gamma {)}^{\mathrm{\prime }}=dS(\dot{\gamma })=\mu \omega (\dot{\gamma })=0$ on the interior of each smooth piece. Thus every horizontal tangent lies in $\ker dS=\ker \omega$ along that piece.

On ${\mathbb{R}}^3$ with coordinates $(x,y,z)$ take $\omega =dz-ydx$. It is smooth and nowhere zero because the $dz$ coefficient is $1$.

Its exterior derivative is $d\omega =dx\wedge dy$, so $\omega \wedge d\omega =(dz-ydx)\wedge dx\wedge dy=dz\wedge dx\wedge dy$, a nonzero coordinate volume form.

If a local $\mu ,S$ as in the refuted claim existed on any open neighborhood, [F2] would force $\omega \wedge d\omega =0$ there, contradicting step 2.1 at every point. Therefore no such local exact multiple exists.

Let $D={\mathbb{R}}^2\setminus \{(0,0)\}$ and $\omega =Pdx+Qdy$ with $P(x,y)=-y\mathrm{/}(x^2+y^2)$ and $Q(x,y)=x\mathrm{/}(x^2+y^2)$. The domain $D$ is open since each nonzero point has a ball avoiding the origin. The coefficient vector $(P,Q)=(-y,x)\mathrm{/}(x^2+y^2)$ is nonzero at every point of $D$. The domain is path-connected: if the segment joining $p,q\in D$ avoids $0$, use that segment; if it contains $0$, then $p,q,0$ are collinear. Writing $p=(p_1,p_2)$, take $r=(-p_2,p_1)$; it is nonzero and perpendicular to $p$, so $p,r$ and $q,r$ are linearly independent pairs and the segments $[p,r]$ and $[r,q]$ avoid $0$.

Write $r^2=x^2+y^2>0$. For sufficiently small $h$, direct difference quotients give ${\mathrm{\partial }}_{x}P=2xy\mathrm{/}{r}^4$, ${\mathrm{\partial }}_{y}Q=-2xy\mathrm{/}{r}^4$, and ${\mathrm{\partial }}_{x}Q={\mathrm{\partial }}_{y}P=(y^2-x^2)\mathrm{/}{r}^4$: for example, $(Q(x+h,y)-Q(x,y))\mathrm{/}h=(y^2-x^2-xh)\mathrm{/}(r^2((x+h{)}^2+y^2))$ and $(P(x,y+h)-P(x,y))\mathrm{/}h=(y^2-x^2+yh)\mathrm{/}(r^2(x^2+(y+h{)}^2))$, while the other two quotients are $y(2x+h)\mathrm{/}(r^2((x+h{)}^2+y^2))$ and $-x(2y+h)\mathrm{/}(r^2(x^2+(y+h{)}^2))$. These limits are continuous on $D$, so $P,Q\in C^1(D)$; the cross partials agree and the coordinate definition [F1] gives $d\omega =({\mathrm{\partial }}_{x}Q-{\mathrm{\partial }}_{y}P)dx\wedge dy=0$.

The unit circle $\gamma (t)=(\cos t,\sin t)$, $0\le t\le 2\pi$, lies in the domain and uses the form defined in step 1.1. Along it $(P,Q)=(-\sin t,\cos t)={\gamma }^{\mathrm{\prime }}(t)$, hence [F2] gives ${\int }_{\gamma }\omega ={\int }_0^{2\pi }1dt=2\pi$ by [F4].

The closedness is proved in step 2.1. If $\omega =dS$ globally on $D$, then $F=\mathrm{\nabla }S$ for the corresponding vector field, and [F3] would give ${\int }_{\gamma }\omega =S(\gamma (2\pi ))-S(\gamma (0))=0$, contradicting step 2.2. Thus $\omega$ is closed but not globally exact. This does not assert that every possible global integrating factor is absent.